Fault Tree Analysis

Part 7 – Common-Mode Failures

THE COMMON–MODE FAILURES WITHIN FAULT TREES

S

1

2

3

SWITCH

PUMP 2

(STAND – BY)POWER 2

POWER 1

PUMP 1

(RUNNING)

Independent Power Source

PUMP 2

SPEED

S

PUMP 1

SPEED PUMP 1

MECH.

FAILURE

POWER 1

FAILURE

PO

WE

R 2

FA

ILU

RE

PU

MP

2 M

EC

H.

FA

ILU

RE

+10 0

10

0

(PU

MP

1 S

PE

ED

= -

10)

-1

SW

ITC

H S

TU

CK

2M

3M

1M

-10

-10

0

1

0

1

0

-10+1

+1

0

-10

+1

+1

0

+10

0

3M ( 10)

AND

G1

OR

P1 Mech Fail.

LocalPower 1Failure

64.97 1041.69 101/ 2.68 yrs.

3.97 hrs.

G2

Pump 1

Shut Down OR

P2MechFail.

LocalPower 2Faiture

SwitchStuck

G3 22.94 10

1/90.8 yrs.

3.95 hrs.

1/3.04 yrs.

4.64 wks.

Pump 2 NotStarted

1 2 3 4 5

41.67 10

1/3 yrs.

4 hrs.

62.3 10

1/ 25 yrs.

5 hrs.

33.8 10

1/5 yrs.

1 wk.

45.5 10

1/35 yrs.

1 wk.

22.5 10

1/10 yrs.

3 mo.

6

4 6 3 6 2

4 3 6 3

4 4 6 4

4 2 6 2

4.97 10

(1.67 10 2.3 10 ) (3.8 10 5.5 10 2.5 10 )

(1.67 10 ) (3.8 10 ) (2.3 10 ) (3.8 10 )

(1.67 10 ) (5.5 10 ) (2.3 10 ) (5.5 10 )

(1.67 10 ) (2.5 10 ) (2.3 10 ) (2.5 10 )

1 3 2 3 1 4 2 4 1 5 2 5PP P P +PP +P P +PP P P

THE COMMON–MODE FAILURES WITHIN FAULT TREES

S

1

2

3

SWITCH

PUMP 2

(STAND – BY)POWER 2

POWER 1

PUMP 1

(RUNNING)

Shared Power Source

PUMP 2

SPEED

S

PUMP 1

SPEED PUMP 1

MECH.

FAILURE

POWER 1

FAILURE

PO

WE

R 2

FA

ILU

RE

PU

MP

2 M

EC

H.

FA

ILU

RE

+10 0

10

0

(PU

MP

1 S

PE

ED

= -

10)

-1

SW

ITC

H S

TU

CK

2M

3M

1M

-10

-10

0

1

0

1

0

-10+1

+1

0

-10

+1

+1

0

+10

PO

WE

R 1

.

FA

ILU

RE

0

0

3M ( 10)

AND

G1

OR

P1 Mech Fail.

LocalPower 1Failure

64.97 1041.69 101/ 2.68 yrs.

3.97 hrs.

G2

Pump 1

Shut Down OR

P2MechFail.

LocalPower 2Faiture

SwitchStuck

G3 22.94 10

1/90.8 yrs.

3.95 hrs.

1/3.04 yrs.

4.64 wks.

Pump 2 NotStarted

1 2 3 4 5

41.67 10

1/3 yrs.

4 hrs.

62.3 10

1/ 25 yrs.

5 hrs.

33.8 10

1/5 yrs.

1 wk.

45.5 10

1/35 yrs.

1 wk.

22.5 10

1/10 yrs.

3 mo.

LocalPower 1Failure

2

GATE MIN CUT SETS

G2

G3

G1

(1) , (2)

(2) , (3) , (4) , (5)

(1 , 2) , (1 , 3) , (1 ,4) , (1 ,5)

(2 , 2) , (2 , 3) , (2 ,4) , (2 , 5)

(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5)

(2) , (2 , 3) , (2 , 4) , (2 , 5)

3M ( 10)

OR

67.23 101/ 20 yrs.

1.27 hrs.

262.3 10 AND

6q 4.2 10

1/120 yrs.

4.4 hrs.

1 541.67 10 22.5 10

AND

76.4 101/ 20 yrs.

4.3 hrs.

1 341.67 10 33.8 10

AND

89.2 101/5333 yrs.

4.6 hrs.

1 441.67 10 45.5 10

COMP q-1(Yr )

1

2

3

4

5

41.67 1062.3 1033.8 1045.5 1022.5 10

1/3

1/25

1/5

1/35

1/10

4 Hr.

5 Hr.

1 Week

1 Week

3 Months

Cut Set KQ K

(2)

(1 , 3)

(1 , 4)

(1 , 5)

6

7

8

6

2.3 10

6.4 10

9.2 10

4.2 10

1 / 25 Yr.

1 / 762 Yr.

1 / 5333 Yr.

1 / 120 Yr.

TOP Event6

TOPQ 7.23 10

TOP1/ 20Yr.

UnavailabilityImportances

Q 6 6

(2)

Q

(1,3)

Q

(1,4)

Q

(1,5)

I 2.3 10 / 7.23 10 .32

I .088

I .0012

I .58

Q

1

Q

2

Q

3

Q

4

Q

5

I .669

I .32

I .088

I .0012

I .53

Unreliability Importances

R

(2)

R

(1,3)

R

(1,4)

R

(1,5)

1/ 25I 0.8

1/ 20I .026

I .004

I 0.167

R

1

R

2

R

3

R

4

R

5

I .197

I 0.8

I .026

I .004

I .167

DEFINITIONSCommon CauseA condition or event which causes multiple basic events in a fault tree.

Common-Mode Event

The basic events caused by the same common cause are called common–mode events of the cause.

Neutral EventFor a given common cause, a basic event is called a neutral event if it is independent of the cause.

Common–Mode Cut SetA cut set is called common–mode cut set when a common cause results in the co–occurrence of all events in the cut set.

CATEGORIES AND EXAMPLES OF COMMON CAUSES

Source Symbol Category Examples

Environment,

System Components,

or Subsystems

I Impact Pipe whip, water hammer, missiles, earthquake, structural failure

V Vibration Machinery in motion, earthquake P Pressure Explosion, out-of –tolerances system changes (pump overspeed,

flow blockage)

G Grit Airborne dust, metal fragments generated by moving parts with inadequate tolerances

S Stress Thermal stress at welds of dissimilar metals, thermal stresses and bending moments caused by high conductivity and density

T Temperature Fire, lightning, welding equipment, cooling system faults, electrical short circuits

E Loss of energy Common drive shaft, same power supply source

C Calibration Misprinted calibration instruction

F Manufacturer Repeated fabrication error, such as neglect to properly coat relay contacts. Poor workmanship. Damage during transportation.

Plant PersonnelIN Installation Same subcontractor or crew

M Maintenance Incorrect procedure, inadequately trained personnel

O Operator or operation Operator disabled or overstressed, faulty operating procedures

TS Test procedure Faulty test procedures which may affect all component normally tested together

Aging A Aging Components of same materials

Top event

2

1

4

5

6 3 12

8

2

7

6

2

5 10 9

2 11

3

1 4

2-out-of-3

Gate Y

2-out-of-3

Gate X

1

4 6 4 6 3 12 3 12

1

2-out-of-3 gate X

6

5 10 5 10 9

2

9

2

6

11 11

2-out-of-3 gate Y

102

5 6

104

3 4 12

199

9

101

1

10

106103

11

105 2

7 8

Common Causes, Domains, And Common–Mode Events Of Example

Category Common Cause Domain Common-Mode Events

ImpactI1I2I3

102, 104101, 103, 105 106

6, 31, 2, 7, 810

StressS1S2S3

103, 105, 106 199101, 102, 104

11, 2, 7, 1091, 4

TemperatureT1T2

106101, 102, 103,104, 105, 199

105, 11, 8, 12, 3, 4

VibrationV1V2

102, 104, 106,101, 103, 105, 199

5, 6, 107, 8

OperatorO1O2

AIIAII

1, 3, 125, 7, 10

Energy source

E1E2

AIIAII

2, 91, 12

Manufacturer F1 AII 2, 11

Installation

Contractor

IN1IN2IN3

AIIAIIAII

1, 126, 7, 103, 4, 5, 8, 9, 11

Test

procedure

TS1TS2

AIIAII

2, 114, 8

SHORT-CUT METHOD FOR OBTAINING COMMON– MODE CUT SETS IN A LARGE FAULT TREE

( )(1) (2)

MIN MIN MIN

TOP EVENT= CUT 1 CUT 2 ...... CUT ...

SET SET SET

MIN

where CUT A A ... A

SET

ni

i i i

i

i

If all the basic events in the ith set are the common-mode events of a common cause, then

jC

r r

MIN

P CUT P C

SETj

i

Otherwise,

( ) ( )1 2

r r

MIN

P CUT P C A A ...

SET

0

j j

j i ii

( Since multiple events appear in the set )

[EXAMPLE]( i ) common cause : O1

common-mode events : 1 , 3 , 12

neutral events : 2 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11

Gate X :

OR

N AND AND

1 3 12

N OR

3 2

AND

1 3 12

Gate Y :

OR

N AND AND

N OR

N N

OR N

N N

OR

N

T

AND

OR OR

AND OR

1 3 12 N AND

N N

N OR

3 1

T

AND

AND OR

1 3 12 1 3

Minimum cut set is { 1 , 3 , 12 }

{ 1 , 3 , 12 } is the common-mode cut set for the common cause O1.

( ii ) common cause : I3

common – mode events : 10

neutral events : 1 , 2 , ……, 9 , 11 , 12

There is no common – mode cut set for I3.

T

X

( iii ) In summary

Common Cause Common – mode Cut Set

I2

I2

S3

S1

T2

O1

{ 1 , 2 }

{ 1 , 7 , 8 }

{ 1 , 4 }

{ 2 , 10 , 11 }

{ 3 , 4 , 12 }

{ 1 , 3 , 12 }

CHAPTER FOUR

Probability Calculation

或然率之計算

RESULTS OF PROBABILITY

CALCULATIONS

1) The probability of the top event.

2) The “importance” of the cut sets and primal events.

概率計算之結果

PROBABILITY OF EVENTS CONNECTED

BY AN “AND” GATE

In general, if events X and Y are probabilistically dependent, then

P(X Y)=P(X Y)P(Y)

Where, is the probability that X occurs given that Y occurs.

If events X and Y are probabilistically independent, then

P(X Y)

P(X Y)=P(X)

P(X Y)=P(X)P(Y)and

Usually, it is assumed that the basic events in a fault tree are independent.

Thus,1 2 1 2

P(B B B ) P(B )P(B ) P(B )n n

PROBABILITY OF EVENTS CONNECTED

BY AN “OR” GATE

1 2 1 2

1 2

P(B B B ) 1 P(B B B )

1 [1 P(B )][1 P(B )] [1 P(B )]n n

n

2,n1 2 1 2 1 2

P(B B ) P(B ) P(B ) P(B )P(B )

3,n1 2 3 1 2 3

1 2 2 3 3 1

1 2 3

P(B B B ) P(B ) P(B ) P(B )

P(B )P(B ) P(B )P(B ) P(B )P(B )

P(B )P(B )P(B )

Note,

1 2 1 2 1 2P(B . .B ) P(B ) P(B ) 2P(B )P(B )EOR

PROBABILITY OF EVENTS CONNECTED

BY A m OUT OF n VOTING GATE

Assume

1 2P(B ) P(B ) P(B ) Q

n

thenkQ Q (1 Q)

nn k

systemk m

n

k

! =

!( )!

n n

k k n k

where

TRANSITION OF COMPONENT STATES

Normal

State

Failed

State

Component

fails

Component

is

repaired

Normal

State

continues

Failed

State

continues

TABLE 4.1 MORTALITY DATA

t = age in year; L(t) = number of living at age t

t L(t)

0

1

2

3

4

5

10

1,023,102

1,000,000

994,230

990,114

986,767

983,817

971,804

t

15

20

25

30

35

40

45

L(t)

962,270

951,483

939,197

924,609

906,554

883,342

852,554

t

50

55

60

65

70

75

80

L(t)

810,900

754,191

677,771

577,822

454,548

315,982

181,765

t L(t)

85

90

95

99

78,221

21,577

3,011

125

TABLE 4.2

t Age in Years

L(t) Number Living atAge t

L(t)R(t)=

NF(t)=1-R(t)

0 1 2 3 4 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 99100

1,023,1021,000,000 994,230 990,114 986,767 983,817 971,804 962,270 951,483 939,197 924,609 906,554 883,342 852,554 810,900 754,191 677,771 577,882 454,548 315,982 181,765 78,221 21,577 3,011 125 0

1.0.97740.97180.96780.96450.96160.94990.94050.93000.91800.90370.88610.86340.83330.79260.73720.66250.56480.44430.30880.17770.07650.02110.00290.0001 .0

0.0.02260.02820.03220.03550.03840.05010.05950.07000.08200.09630.11390.13660.16670.20740.26280.33750.43520.55570.69120.82230.92350.97890.99710.99991.

0 10 20 30 40 50 60 70 80 90 100

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

PSurvival distribution

Failur

e dist

ributio

n

Pro

babi

lity

of

Sur

viva

l R(t

) an

d D

eath

F(t

)

TABLE 4.3 FAILURE DENSITY FUNCTION f(t)

Age in Years ( ) ( )n t n t

No. of Failures(death)

2

( ) ( )( )

n t n tf t

N

2

( )( )

dF tf t

dt

0 1 2 3 4 510 152025303540455055606570758085909599100

23,102 5,770 4,116 3,347 2,950 12,013 9,534 10,787 12,286 14,588 18,055 23,212 30,788 41,654 56,709 76,420 99,889123,334138,566134,217103,554 56,634 18,566 2,886 125 0

0.022600.005640.004020.003270.002880.002350.001860.002110.002400.002850.003530.004540.006020.008140.011100.015000.019500.024100.027100.026200.020200.011100.003630.000710.00012

0.005400.004540.002840.003300.002870.001920.001980.002240.002590.003640.003930.004360.006370.009620.013670.018000.022000.024900.024600.019500.009700.00210 _- - -

20 40 60

20

40

60

80

80

100

120

140

100

Num

ber

of D

eath

s (t

hous

ands

)

Age in Years (t)

20 40 60

0.2

0.4

0.6

80

0.8

0.10

0.12

0.14

100 Age in Years (t)

Fai

lure

Den

sity

f (

t)

THE REPAIR - TO - FAILURE PROCESSES

Reliability, R(t)*

The probability that the component experiences no failure during the time interval (0, t), given that

the component was repaired at time zero.0

lim ( ) 1 lim ( ) 0t t

R t R t

Unreliability, F(t)*The probability that the component experiences the first failure during (0, t), given that it is repaired at time zero. Note, R(t) + F(t) = 1.

0lim ( ) 0 lim ( ) 1

t tF t F t

Failure Density, f(t)*

The probability that the component experiences a failure per unit time at time t, given that the component was repaired at time zero. ( )

( )dF t

f tdt

Note,

( ) ( ) ( ) ( )0

tF t f u du R t f u du

t

Failure Rate, r(t)*

The probability that the component experiences a failure per unit time at time t, given that

the component was repaired at time zero and has survival to time t.

( ) ( )( )

( ) 1 ( )

f t f tr t

R t F t

Mean Time to Failure, MTTF

( ) 0

MTTF tf t dt

R(t)+F(t) = 1

The component with a constant failure rate, r, is considered as good as new, if it is functioning.

t-dR/dtr(t) = R(t) = exp [- ( ) ]

0Rr u du

number of deaths during [t, t+ ] f (t)r(t)

number of survivals at age t R(t)

f(t) f(t)r(t)=

R(t) 1-F(t)

Age in

Years

No. of Failures

(death)

( )( )

1 ( )

f tr t

F t

0

1

2

3

4

5

10

15

20

25

30

35

23,102

5,770

4,116

3,347

2,950

12,013

9,543

10,787

12,286

14,588

18,055

23,212

0.02260

0.00570

0.00414

0.00338

0.00299

0.00244

0.00196

0.00224

0.00258

0.00311

0.00391

0.00512

Age in

Years

40

45

50

55

60

65

70

75

80

85

90

95

99

No. of Failures

(death)

30,788

41,654

56,709

76,420

99,889

123,334

138,566

134,217

103,554

56,634

18,566

2,886

125

( )( )

1 ( )

f tr t

F t

0.00697

0.00977

0.01400

0.02030

0.02950

0.04270

0.06100

0.08500

0.11400

0.14480

0.17200

0.24000

1.20000

20 40 60 80 100

0.05

0.1

0.15

0.2Early failures

Random failures

Wearout failuresF

ailu

re R

ate

r(t)

Failure rate r(t) versus t.

THE FAILURE-TO-REPAIR PROCESSES

Repair Probability, G(t) F(t)

The probability that repair is completed before time t, given that the component failed at time zero.

Note, G(t)=0 if the component is non-repairable.

Repair Density, g(t)( )

( )dG t

g tdt

Note, ( ) ( )

0

tG t g u du

Repair Rate, m(t)*

The probability that the components repaired per unit time at time t, given that the component failed at time zero and has been failed to time t. Note, m(t)=0 if the component is non-repairable.

Mean Time to Repair, MTTRM(t)=g(t)/(1-G(t))

( )

0MTTR tg t dt

* A component with a const repair rate has the same chance of being repaired whenever it is failed.

THE WHOLE PROCESS

[ EXAMPLE ]

6A(5)= 0.6

10Q(5)=1-A(5)=0.4

2 3(5) 0.2 (6) 0.3

10 100 3

(5) 0 (6) 0.310 102

(5) 0.3336

(5) 0u

6 (0,6) (t)dt

0

(0) 1 (1) 1 (5) 1

W

Availability, A(t) : The probability of a component being normal at t.

Unavailability, Q(t) : The probability of a component being failed at t.

Unconditional Failure Intensity, W(t) : The probability that a component fails per fails per

unit time at t.

Conditional Failure Intensity, : The proportion of the normal population that is expected

to fail per unit time at t.

( )t

Expected Number of Failures, 1 2

( , )W t t

2

1 2

1

( , ) ( )

tW t t t dt

t